Chapter 27, Problem 72GP

(a)

To determine

Potential difference for a proton.

Answer to Problem 72GP

  $33V$ .

Explanation of Solution

Given:

Mass is $1.67\times {10}^{-27}kg$

Wavelength is $5.0\times {10}^{-12}m$

Formula used:

Each absorbed photon’s momentum changes form $\text{p}\text{=}\frac{\text{h}}{\text{λ}}$ or $\text{pc}\text{=}\frac{\text{hc}}{\text{λ}}$

Where,

  $\lambda$ is wavelength.

  $p$ is photon’s momentum

  $h$ is Planck’s constant, $6.63\times {10}^{-34}J\cdot s$

  $c$ is speed of light, $3.00\times {10}^{8}m/s$

It is known that kinetic energy: $\text{KE=}\frac{{\text{p}}^{\text{2}}}{{\text{2m}}_{\text{0}}}\text{=}\frac{{\left(\text{pc}\right)}^{\text{2}}}{{\text{2m}}_{\text{0}}{\text{c}}^{\text{2}}}$

Where,

  $p$ is photon’s momentum

  $c$ is speed of light, $3.00\times {10}^{8}m/s$

  $m_0$ is rest mass

It is known that potential difference: $\text{V=}\frac{\text{KE}}{\text{e}}$

Where,

  $e$ is electron

  $KE$ is kinetic energy

Calculation:

The required momentum is

  $pc=\frac{hc}{\lambda }=\frac{\left(1.24\times {10}^{3}eV\cdot nm\right)}{5.0\times {10}^{-3}nm}=2.48\times {10}^{5}eV=0.248MeV$

For the proton, $pc<<m_0{c}^2$ , so, calculate the required kinetic energy from

  $\text{KE=}\frac{{\text{p}}^{\text{2}}}{{\text{2m}}_{\text{0}}}\text{=}\frac{{\left(\text{pc}\right)}^{\text{2}}}{{\text{2m}}_{\text{0}}{\text{c}}^{\text{2}}}$

  $\text{=}\frac{{\left(\text{0}\text{.248}\text{MeV}\right)}^{\text{2}}}{\text{2}\left(\text{938}\text{MeV}\right)}\text{=3}{\text{.3×10}}^{\text{-5}}\text{MeV=33}\text{eV}$

The potential difference to produce this kinetic energy is $\text{V=}\frac{\text{KE}}{\text{e}}$

  $\text{V=}\frac{\left(\text{33}\text{eV}\right)}{\text{1e}}\text{=33}\text{V}$

(b)

To determine

Potential difference for an electron.

Answer to Problem 72GP

  $\text{57}\text{kV}$ .

Explanation of Solution

Given:

Mass is $9.11\times {10}^{-31}kg$

Wavelength is $5.0\times {10}^{-12}m$

Formula used:

It is known that kinetic energy: $\text{KE=}\frac{{\text{p}}^{\text{2}}}{{\text{2m}}_{\text{0}}}\text{=}\frac{{\left(\text{pc}\right)}^{\text{2}}}{{\text{2m}}_{\text{0}}{\text{c}}^{\text{2}}}$

Where,

  $p$ is photon’s momentum

  $c$ is speed of light, $3.00\times {10}^{8}m/s$

  $m_0$ is rest mass

It is known that potential difference: $\text{V=}\frac{\text{KE}}{\text{e}}$

Where,

  $e$ is electron

  $KE$ is kinetic energy

Calculation:

For the electron, $pc$ is on the order of $m_0{c}^2$ .

Thus, calculate the required kinetic energy from

  $\text{KE=}{\left[{\left(\text{pc}\right)}^{\text{2}}\text{+}{\left({\text{m}}_{\text{0}}{\text{c}}^{\text{2}}\right)}^{\text{2}}\right]}^{\frac{\text{1}}{\text{2}}}{\text{-m}}_{\text{0}}{\text{c}}^{\text{2}}$

  $\text{KE=}{\left[{\left(\text{0}\text{.248}\text{MeV}\right)}^{\text{2}}\text{+}{\left(\text{0}\text{.511}\text{MeV}\right)}^{\text{2}}\right]}^{\frac{\text{1}}{\text{2}}}\text{-0}\text{.511}\text{MeV=0}\text{.057}\text{MeV=57}\text{}\text{KeV}$

The potential difference to produce this kinetic energy is $\text{V=}\frac{\text{KE}}{\text{e}}$

  $\text{V=}\frac{\left(\text{57}\text{KeV}\right)}{\text{1e}}\text{=57}\text{kV}$